1. ## Equation system excercise

Solve for every a the equationsystem:

$\displaystyle x+y+az=3$
$\displaystyle 2x-y+3x=4$
$\displaystyle x+ay+z=3$

I put the system in a matrix and did elimination until:

1 2 a 3
0 -3 (3-2a) -2
0 (a-1) (1-a) 0

which gives me a = (y-z)/(y-z)=1

Is this correct, I'm not very used to this kind of exercises with unknown coefficients so I'm rather unsure at this. Am I going in the right direction? If not I'd be grateful for any help!

2. No one can help me?

3. Originally Posted by Boneleg
Solve for every a the equationsystem:

x+y+az=3

2x-y+3z=4 <<<<<<<<<< typo?

x+ay+z=3

I put the system in a matrix and did elimination until:

1 2 a 3
0 -3 (3-2a) -2
0 (a-1) (1-a) 0

which gives me a = (y-z)/(y-z)=1

Is this correct, I'm not very used to this kind of exercises with unknown coefficients so I'm rather unsure at this. Am I going in the right direction? If not I'd be grateful for any help!
1. I assume that there is a typo at the indicated line(?)

2. In my opinion you have to solve this system for (x, y, z) with respect to a.

3. I used determinants and I got $\displaystyle (x, y, z)=\left(2-\dfrac1a\ ,\ \dfrac1a\ ,\ \dfrac1a \right)~,~a \neq 0$